This invention relates to circuits that generate electrical currents proportional to an exponential function of one or more input currents.
If I.sub.d is the current flowing through a diode, then the voltage across the diode is equal to V.sub.t .multidot.1n[I.sub.d +I.sub.s)/I.sub.s], where I.sub.s is the saturation current of the diode. V.sub.t =K.sub.B T/q, where K.sub.B is Boltzmann's constant, T is the temperature, and q is the charge of an electron. Since I.sub.s is typically in the range of 10.sup.-18 to 10.sup.-16 amperes, and I.sub.d &gt;&gt;I.sub.s, the voltage across the diode closely approximates V.sub.t .multidot.1n(I.sub.d /I.sub.s). Likewise, the voltage across the base-emitter junction of a transistor closely approximates V.sub.t .multidot.1n(I.sub.d /I.sub.s) where I.sub.c is the current flowing into the collector of the transistor.
FIG. 1 shows a circuit 100 that produces an output current I.sub.o equal to the square root of the product of currents I.sub.1 and I.sub.2. The saturation current I.sub.s is the same for all of the transistors in the circuit. Current source 102 produces current I.sub.1 and current source 104 produces current I.sub.2. Current source 102 is connected between a voltage source 106 and the collector of transistor 108. The emitter of transistor 108 is connected to ground. The voltage at the base of transistor 108 is therefore V.sub.t .multidot.1n(I.sub.1 /I.sub.s). The base of transistor 108 is connected to the emitter of transistor 110. Current source 104 is connected between the emitter of transistor 110 and ground. The collector of transistor 110 is connected to the voltage source 106. The voltage at the base of transistor 110 is therefore V.sub.t .multidot.1n(I.sub.1 /I.sub.s)+V.sub.t .multidot.1n(I.sub.2 /I.sub.s). The base of transistor 110 is connected to current source 102 and the base of transistor 112. The emitter of transistor 112 is connected to the collector and base of transistor 114, which functions as a diode. The emitter of transistor 114 is connected to ground. The voltage at the base of transistor 112 is therefore 2V.sub.t .multidot.1n(I.sub.o /I.sub.s). Thus, EQU 1n(I.sub.1 /I.sub.s)+1n(I.sub.2 /I.sub.s)=21n(I.sub.o /I.sub.s), or I.sub.o =(I.sub.1 .multidot.I.sub.2)1/2.
Other circuits produce an output voltage equal to the square root of an input voltage. For example an operational amplifier can be connected with a diode in its feedback loop, so that the operational amplifier produces an output proportional to the logarithm of an input voltage. The logarithm output is connected to a voltage divider that produces an output voltage equal to one-half of the input voltage to the voltage divider. The output of the voltage divider is connected to the inverting input of a second operational amplifier through a diode, so that the second amplifier produces an output proportional to the antilogarithm of the output of the voltage divider. Thus, EQU log(V.sub.out)=1/2[log(V.sub.in)], or V.sub.out =V.sub.in 1/2.
In another circuit, an input voltage V.sub.in is connected through a resistor to the inverting input of an operational amplifier. The output, V.sub.out, of the operational amplifier is connected to a multiplier circuit whose output is equal to -(V.sub.out).sup.2. The output of the multiplier circuit is connected through a resistor to the inverting input of the operational amplifier. V.sub.out equals V.sub.in 1/2.